$\sum\limits_{n=1}^{\infty } \dfrac{(x+2)^n}{(n+1)\cdot(-2)^n} $ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $-1 < x \le 3$ (Choice B) B $-4<x \le 0$ (Choice C) C $-4 \le x < 0$ (Choice D) D $-1 \le x < 3$
We use the ratio test. For $x\neq -2$, let $a_n= \dfrac{(x+2)^n}{(n+1)\cdot(-2)^n} $. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x+2}{2}\right| $ The series converges when $\left|\dfrac{x+2}{2}\right|<1$, which is when $-4<x<0 $. Now let's check the endpoints, $x=-4$ and $x=0$. Letting $x=-4$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(-4+2)^n}{(n+1)\cdot(-2)^n} &=\sum\limits_{n=1}^{\infty } \dfrac{(-2)^n}{(n+1)\cdot(-2)^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{1}{n+1} \end{aligned}$ By the integral test, we know this series diverges. Letting $x=0$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(0+2)^n}{(n+1)\cdot(-2)^n} &=\sum\limits_{n=1}^{\infty }\dfrac{2^n}{(n+1)\cdot(-2)^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n \cdot2^n}{(n+1)\cdot2^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n}{n+1} \end{aligned}$ By the alternating series test, we know this series converges. In conclusion, the interval of convergence is $-4<x \le 0$.